A321477 -id:A321477 - cp's OEIS frontend

A323017 T(n,k) = A321477(n,k)/A321476(n,k), 0 <= k <= n - 1.

1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 4, 4, 4, 4, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 4, 2, 4, 2, 1, 2, 4, 2, 4, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2

Offset: 1

Jianing Song, Jan 07 2019

For Lucas sequences, say, rows in A172236, we are mainly concerned about the periods, ranks and the ratios of the periods to the ranks of them modulo a given integer n. The period of {A172236(k,m) modulo m} is given as A321477(n,k), and the rank, which is defined as the smallest l > 0 such that n divides A172236(k,l), is given as A321476(n,k). T(n,k) is their ratio, which is the multiplicative order of A172236(k,A321476(n,k)+1) modulo n.
T(n,k) has value 1, 2 or 4. This is because A172236(k,m+1)^4 == 1 (mod A172236(k,m)). For n > 2, T(n,k) = 4 iff A321476(n,k) is odd; 1 iff A321476(n,k) is even but not divisible by 4; 2 iff A321476(n,k) is divisible by 4. See A172236 for some further properties.
Let p be an odd prime. If p == 3 (mod 4), the p^e-th row consists of only 1 and 2; if p == 5 (mod 8), the p^e-th row consists of only 1 and 4.

			Table begins
  1,
  1, 1,
  1, 2, 2,
  1, 1, 1, 1,
  1, 4, 4, 4, 4,
  1, 2, 2, 1, 2, 2,
  1, 2, 1, 2, 2, 1, 2,
  1, 2, 1, 2, 1, 2, 1, 2,
  1, 2, 2, 1, 2, 2, 1, 2, 2,
  1, 4, 2, 4, 2, 1, 2, 4, 2, 4,
  ...

Cf. A172236, A321476, A321477.

A172236(k, m) = ([k, 1; 1, 0]^m)[2, 1]
T(n, k) = my(i=1); while(A172236(k, i)%n!=0, i++); znorder(Mod(A172236(k, i+1), n))

A321476 Regular triangle read by rows: T(n,k) is the rank of {A172236(k,m)} modulo n, 0 <= k <= n - 1.

Original entry on oeis.org

1, 2, 3, 2, 4, 4, 2, 6, 4, 6, 2, 5, 3, 3, 5, 2, 12, 4, 6, 4, 12, 2, 8, 6, 8, 8, 6, 8, 2, 6, 8, 6, 4, 6, 8, 6, 2, 12, 12, 6, 4, 4, 6, 12, 12, 2, 15, 6, 3, 10, 6, 10, 3, 6, 15, 2, 10, 12, 4, 10, 12, 12, 10, 4, 12, 10, 2, 12, 4, 6, 4, 12, 4, 12, 4, 6, 4, 12

Offset: 1

Jianing Song, Nov 11 2018

The rank of {A172236(k,m)} modulo n is the smallest l such that n divides A172236(k,l).
Though {A172236(0,m)} is not defined, it can be understood as the sequence 0, 1, 0, 1, ... So the first column of each row (apart from the first one) is always 2.
Every row excluding the first term is antisymmetric, that is, T(n,k) = T(n,n-k) for 1 <= k <= n - 1.
T(n,k) is the multiplicative order of -((k + sqrt(k^2 + 4))/2)^2 modulo n*sqrt(k^2 + 4), where the multiplicative order of u modulo z is the smallest positive integer l such that (u^l - 1)/z is an algebraic integer.

			Table begins
  1;
  2,  3;
  2,  4,  4;
  2,  6,  4,  6;
  2,  5,  3,  3,  5;
  2, 12,  4,  6,  4, 12;
  2,  8,  6,  8,  8,  6,  8;
  2,  6,  8,  6,  4,  6,  8,  6;
  2, 12, 12,  6,  4,  4,  6, 12, 12;
  2, 15,  6,  3, 10,  6, 10,  3,  6, 15;
  ...

Cf. A172236, A321477 (periods).

A172236(k, m) = ([k, 1; 1, 0]^m)[2, 1]
T(n, k) = my(i=1); while(A172236(k, i)%n!=0, i++); i

Let p be an odd prime. (i) If k^2 + 4 is not divisible by p: if p == 1 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - ((k^2+4)/p))/2; if p == 3 (mod 4), then T(p^e,k) is divisible by p^(e-1)*(p - ((k^2+4)/p)) but not divisible by p^(e-1)*(p - ((k^2+4)/p))/2. Here (a/p) is the Legendre symbol. (ii) If k^2 + 4 is divisible by p, then T(p^e,k) = p^e.
For e >= 3 and k > 0, T(2^e,k) = 3*2^(e-2) for odd k and 2^(e-v(k,2)+1) for even k, where v(k,2) is the 2-adic valuation of k.
If gcd(n_1,n_2) = 1, then T(n_1*n_2,k) = lcm(T(n_1,k mod n_1),T(n_2, k mod n_2)).
T(n,k) <= 2*n.

cp's OEIS Frontend

A323017 T(n,k) = A321477(n,k)/A321476(n,k), 0 <= k <= n - 1.

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